1. The problTfem statement, all variables and given/known data Determine the result when 100g of steam at 100C is passed into a mixture of 200g of Water and 20g of ice at exactly 0C in a calorimeter which behaves thermally as if it were equvalent to 30g of water. 2. Relevant equations FInding the mass. I think I figured out the final temp? 3. The attempt at a solution Q=0=Steam condensed heat + Change in steam water heat + change in water heat + heat to melt Ice + change in calorimeter heat (540cal/g)(100g) + (100g)(1cal/g C)(Tf-0C) +(200g)(1cal/g C)(Tf-0C) +(20g)(80g/C) +(20g)(1cal/g C)(Tf-0C) + (30g)(1cal/gC)(Tf-0C) Tf=149.7 ITs all water so the temp has to be 100 it can't go higher than that. I am struggling to find the grams of steam condensed. m(-540g/C)=Ice melted + Rise in water? Please help and thanks!